How to capture the exception raised in the scriptblock of start-job?

Question

I have the following script,

$createZip = {
    Param ([String]$source, [String]$zipfile)
    Process { 
        echo \"zip: $source`n     --> $zipfile\"
         


        

Answer 1

Using throw will change the job object's State property to "Failed". The key is to use the job object returned from Start-Job or Get-Job and check the State property. You can then access the exception message from the job object itself.

Per your request I updated the example to also include concurrency.

$createZip = {
    Param ( [String] $source, [String] $zipfile )

    if ($source -eq "b") {
        throw "Failed to create $zipfile"
    } else {
        return "Successfully created $zipfile"
    }
}

$jobs = @()
$sources = "a", "b", "c"

foreach ($source in $sources) {
    $jobs += Start-Job -ScriptBlock $createZip -ArgumentList $source, "${source}.zip"
}

Wait-Job -Job $jobs | Out-Null

foreach ($job in $jobs) {
    if ($job.State -eq 'Failed') {
        Write-Host ($job.ChildJobs[0].JobStateInfo.Reason.Message) -ForegroundColor Red
    } else {
        Write-Host (Receive-Job $job) -ForegroundColor Green 
    }
}